the rate constant of a reaction …
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Gaurav Seth 5 years, 8 months ago
Relation between equilibrium constant k, Ea and T is given by the equation
ln k = – Ea/RT + ln A
If we have different rate constant at different temperature then given equation can be modified to find out the Ea and A as follows:
log k2/log k1 = Ea/2.303 R[T2-T1/T1T2]...(1)
Putting the values in equation (1) we get,
log 0.07/log 0.02 = Ea/2.303 x 8.314 J K-1 mol-1 (700-500/700 x 500)
0.544 = Ea × 5.714 × 10-4 /19.15
Ea = 0.544 × 19.15 / 5.714 × 10–4 = 18230.8 J
Since k = Ae-Ea/RT
Here we can take any value of k and its corresponding temperature
0.02 = Ae-18230.8/8.314 × 500
(Note to calculate e-18230.8/8.314 × 500 you can find the antilog of total solved value
antilog of -18230.8/8.314 × 500 is 0.012)
A = 0.02/0.012 = 1.61
0Thank You